Question 1055003: Doctors nationally believe that 68% of a certain type of operation are successful. In a particular hospital, 47 of these operations were observed and 37 of them were successful. At = .05, is this hospital's success rate different from the national average?
A) No, because the test value 0.93 is in the noncritical region.
B) Yes, because the test value 0.93 is in the critical region.
C) No, because the test value 1.58 is in the noncritical region.
D) Yes, because the test value 1.58 is in the noncritical region.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Ho: p = 0.68 (claim)
Ha: p ≠ 0.68
sample proportion = p-hat = 37/47 = 0.7872
---
test stat:: z(0.7872) = (0.7872-0.68)/sqrt[0.68*0.32/47] = 1.58
-----
At = .05, is this hospital's success rate different from the national average
C) No, because the test value 1.58 is in the noncritical region ( we do not reject Ho)
p-value = P(z > 1.58) = 0.057
|
|
|
